Velocity versus time form acceleration versus time

[KillerZ]

Homework Statement

I have some data from a lab conducted with a accelerometer to collect the acceleration and I am trying to figure out how to create a v vs t graph from the a vs t graph. When I create the a vs t graph in excel its very crazy looking because of the change in acceleration at every 0.1 s from 0 s to 79.6 s. I know the v vs t is constructed from the area under the a vs t graph. The speed of the object is 0 km/h to 40 km/h = 0 m/s to 11.1111 m/s. I have every point of the instantaneous acceleration at every 0.1 s.

my a vs t graph:

my acceleration data looks like this from 0 s to 79.6 s:

Homework Equations

$$\Delta v = \int a dt$$

The Attempt at a Solution

well I know the time is 0 s to 79.6 s and the velocity is 0 m/s to 11.1111 m/s so I think I can integrate?

$$\Delta v = \int a dt$$

 

[kuruman]

Integration is actually a discrete sum in this case. You don't have an analytical function that you can integrate. Just multiply the acceleration at a given point in time by the time interval Δt = 0.1 s and call that Δv. Add all the Δv's from time zero to the time of interest to get the velocity at the time of interest.

 

[KillerZ]

I used this formula:

$$v_{n} = v_{n-1} + a_{n}(0.1)$$

which is what you said but would the position be:

$$x_{n} = x_{n-1} + v_{n-1}(0.1) + (0.5)a_{n}(0.1)^{2}$$

$$y_{n} = y_{n-1} + v_{n-1}(0.1) + (0.5)a_{n}(0.1)^{2}$$

for every point?

 

[kuruman]

It will (probably) be better to use the "other kinematic equation"

$$\Delta x_{n} = \frac{v_{n}^2-v_{n-1}^2}{2a_{n}}$$